Does the inclusion map satisfy Unique (CP) Extension Property? Consider the $C^\star$ algebra $C(\mathbb{T}^2)$ of complex-valued continuous functions on the 2-torus with the sup norm. Let $S$ be the operator system that is the (norm) closure of the linear span of $\{z^m w^n| m,n \in \mathbb{Z}, mn \geq 0\}$. Let $i:S\to C(\mathbb{T}^2)$ denote the (canonical) inclusion map of $S$ in $C(\mathbb{T}^2)$. $i$ is completely positive (CP), and by Arveson's Extension Theorem, it must have CP extension(s) $\tilde{i}: C(\mathbb{T}^2) \to C(\mathbb{T}^2)$, such that $\tilde{i}|_S=i$. One such CP extension is, $\tilde{i}=\mathbb{I}_{C(\mathbb{T}^2)}$, which is, extending as the identity map on $C(\mathbb{T}^2)$. My question is, is it the only CP extension? Or, in other words, does the inclusion map have the unique extension property (CP) in this case? If not, what would be another example of a CP extension?
Perhaps an interesting point to note: the trivial extension doesn't work, as aptly demonstrated in the answer to a
previous question here.
 A: Yes,  the unique CP extension applies to the present situation.  It is in fact a consequence of the following more
general fact:
Theorem. Let $A$ be a unital C$^*$-algebra and let $\mathscr U$ be a set of unitaries in $A$,
generating $A$ as a C$^*$-algebra,  and  such that $1\in \mathscr U$.
Given a unital representation $\pi :A\to B(H)$, suppose that $\varphi :A\to B(H)$ is a completely positive map such that
$$
  \varphi (u)=\pi (u), \quad\forall u\in \mathscr U.
  \tag 1
  $$
Then $\varphi =\pi $.
Proof.  By Stinespring Theorem,  there exists a Hilbert space $\tilde H$, a  unital representation  $\tilde\pi:A\to B(\tilde H)$,
and a bounded operator $V:H\to \tilde H$, such that
$$
  \varphi (a)=V^*\tilde \pi (a)V, \quad\forall a\in A.
  \tag 2
  $$
Observing that both $\varphi $ and $\pi $ are unital maps, we deduce that $V^*V=1$, which is to say that $V$ is an isometry.
Notice that, for each $u$ in $\mathscr U$ we have
$$
  \pi (u) = \varphi (u) = V^*\tilde \pi (u)V,
  $$
so $\tilde \pi (u)$ commutes with $VV^*$ by Lemma (1), below.
Since the commutant of $VV^*$ is a closed $^*$-subalgebra, and since $A$ is generated as a C$^*$-algebra by $\mathscr U$, it
follows that $\tilde \pi (a)$ commutes with $VV^*$ for every $a$ in $A$.
From (2) we then deduce  that $\varphi $ is a $^*$-homomorphism, and then (1) implies that $\varphi =\pi $.   $ \tag*{$\square$}$

Lemma 1.  If $V:H\to\tilde H$ is an isometry, and $U$ is a unitary operator on $\tilde H$ such that  $V^*UV$
is unitary, then $U$ commutes with $VV^*$.
Proof.  Identifying  $H$ with  a subspace of $\tilde H$ via $V$,  we may write $\tilde H = H \oplus  H^\perp$, and hence $U$ may be
represented as a matrix
$$
  U = \pmatrix {A & B \cr C & D},
  $$
where $A = V^* UV$, hence $A$ is unitary.  It follows that
$$
  \pmatrix {1 & 0 \cr 0 & 1} =   \pmatrix {A & B \cr C & D}   \pmatrix {A^* & C^* \cr B^* & D^*} = \pmatrix {AA^* + BB^*
&\cdots  \cr \cdots & \cdots},
  $$
so $B=0$, and a similar argument proves that $C=0$.  This implies that $U$ commutes with
$$
  \pmatrix {1 & 0 \cr 0 & 0} = VV^*.
  \tag*{$\square$}
  $$
